Question

Q3 A dart is randomly thrown and lands within the boundaries of a 6 foot by 6 foot square. The unshaded regions are each a quarter of an inscribed circle. What is the probability that the dart lands in one of the shaded regions? Express your answer as a common fraction in terms of pi.​ PLEASE GIVE THE FULL EXPLANATION AND NOT JUST THE ANSWER!!!

Answer 1

Answer:

$Pr = \frac{8 - \pi}{8}$

Step-by-step explanation:

Given

See attachment for square

Required

Probability the $dart\ lands$ on the shaded region

First, calculate the area of the square.

$Area = Length^2$

From the attachment, Length = 6;

So:

$A_1 = 6^2$

$A_1 = 36$

Next, calculate the area of the unshaded region.

From the attachment, 2 regions are unshaded. Each of this region is quadrant with equal radius.

When the two quadrants are merged together, they form a semi-circle.

So, the area of the unshaded region is the area of the semicircle.

This is calculated as:

$Area = \frac{1}{2} \pi d$

Where

$d = diameter$

d = Length of the square

$d =6$

So, we have:

$Area = \frac{1}{2} \pi (\frac{d}{2})^2$

$Area = \frac{1}{2} \pi (\frac{6}{2})^2$

$Area = \frac{1}{2} \pi (3)^2$

$Area = \frac{1}{2} \pi *9$

$Area = \pi * 4.5$

$Area = 4.5\pi$

The area (A3) of the shaded region is:

$A_3 = A_1 - A_2$ ---- Complement rule.

$A_3 = 36 - 4.5\pi$

So, the probability that a dart lands on the shaded region is:

$Pr = \frac{A_3}{A_1}$ i.e. Area of shaded region divided by the area of the square

$Pr = \frac{36 - 4.5\pi}{36}$

Factorize:

$Pr = \frac{4.5(8 - \pi)}{36}$

Simplify

$Pr = \frac{8 - \pi}{8}$